Seek Earth's Geography in Alg-Geometry

Let $PB = a \quad , \quad PA = b$ .

Since $a,b>0$ , Using $R.M.S$ inequality,

$\dfrac{a+b}{2} \leq \sqrt{\dfrac{a^2+b^2}{2}} \\ \Rightarrow \dfrac{a+b}{2} \leq \sqrt{\dfrac{(2r)^2}{2}} \\ \Rightarrow \dfrac{a+b}{2} \leq \dfrac{2r\sqrt{2}}{2} \\ \Rightarrow a+b \leq 2r\sqrt{2}$

We see that maximum is attained at equality $a=b=r\sqrt{2}$

This implies that $PA+PB$ is maximum when $\Delta PAB$ is a $45^0-45^0-90^0$ triangle.

$\text{Area of triangle ABC}=\dfrac{1}{2} \times r\sqrt{2} \times r\sqrt{2} = r^2$

$\text{Area of semicircle} = \dfrac{1}{2} \times \dfrac{22}{7} \times r \times r = \dfrac{11}{7}r^2$

$\Rightarrow \text{Required area} = \dfrac{11}{7}r^2 - r^2 = \boxed{\dfrac{4}{7}r^2}$

Substituting $r=6370$ gives us the area as $\Large\boxed{\color{#3D99F6}{23186800}}$

Now that we have an algebraic and a calculus solution, let me suggest a geometric solution as well, for the sake of variety.

Let $t$ be the angle $PAB$ . Then $AP+PB=2r\cos{t}+2r\sin{t}=2\sqrt{2}r\sin(t+\pi/4).$ The maximum is attained when $t=\pi/4$ , making the sine equal to 1. Now the area we seek is $\frac{\pi{r^2}}{2}-\frac{2r\times{r}}{2}\approx\frac{4r^2}{7}=\boxed{23186800}$

It can be a Geometry problem but we must use Calculus.

We know that $\angle APB = 90^\circ$ . If $\angle PAB = \theta$ , then $PA+PB = l = 2r(\cos{\theta} + \sin{\theta})$ .

$\dfrac {dl}{d\theta} = 2r(-\sin{\theta} + \cos{\theta})$ , when $\dfrac {dl}{d\theta} = 0 \quad \Rightarrow \tan{\theta} = 1 \quad \Rightarrow \theta = 45^\circ$ .

We note that $\dfrac {d^2l}{d\theta^2} < 0$ . Therefore, $PA+PB$ is maximum when $\theta = 45^\circ$ and $PA=PB=\sqrt{2}r$ .

The area of $\color{#3D99F6} {blue}$ region

$\begin{aligned} \color{#3D99F6}{A_{blue}} & = \text{Area of semicircle} -\triangle APB \text{ (when } PA=PB=\sqrt{2}r ) \\ & = \dfrac {\pi r^2}{2} - \dfrac {(\sqrt{2}r)^2}{2} = \dfrac {\pi - 2}{2}r^2 = \left( \dfrac {11}{7}-1\right) 6370^2 \\ & = \boxed{23186800} \end{aligned}$ .

Extreme values are obtained in this symmetric conditions when PA=PB. So brown area of isosceles right triangle is $\frac 1 2 *(\sqrt 2*r)^2 , ~~Semicircle~ area =\frac 1 2 *\pi*r^2\\ \therefore~blue ~area=\{\frac{11} 7 ~- ~1\}*6370^2=2318,6800.$ .